===============================================================================
A ZERO-ENTROPY SYSTEM WITHOUT A SMOOTH POSITIVE-VOLUME MODEL
A plain-language rewrite for readers of every age
===============================================================================
Original title: "A zero-entropy system without a smooth positive-volume model"
Original author: OpenAI, September 2026.
This file says the same thing as the original paper. It does not change the
mathematics. It states each definition in words first, then in symbols,
then gives the reason every step of the proof is there. Every Greek letter is
introduced by an English word before it is used. Every named theorem is
stated where it is first needed.
The rewrite is more concise: it does not repeat the same argument twice.
The rewrite is more complete: every lemma used in the final contradiction is
stated in words here, so a reader can follow the chain of logic from the
first sentence of the paper to the last.
===============================================================================
CONTENTS
===============================================================================
0. How to read this file
1. The question
2. Entropy, in one page
3. What the main theorem says
4. What was known before this paper
5. Why positive density is harder than full support
6. The main geometric engine: the switch
7. Stationary estimates: summable dyadic errors
8. Linear estimates: how to split the tangent space
9. Building the switch: cells, joins, Jacobian
10. The symbolic system: four word properties
11. The process: zero entropy, big window entropy
12. The obstruction: upper bound vs. lower bound
13. What this paper does NOT claim
14. Glossary of every symbol
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0. HOW TO READ THIS FILE
===============================================================================
If you have had pre-calculus and know what a probability is, you can read
sections 1 through 5 in one sitting. Sections 6 through 10 are the geometry.
Sections 11 and 12 are combinatorics, which is just careful counting.
Section 13 is the punchline: two numbers that cannot both be true.
Nothing here requires a theorem you have not seen. If a word is unfamiliar,
look in the glossary at the end.
===============================================================================
1. THE QUESTION
===============================================================================
Start with a set X of "states" and a probability measure mu on X. So mu is a
rule that assigns a number between 0 and 1 to every measurable subset, mu(X) = 1,
and mu is countable-additive: if pieces are disjoint, their probabilities add.
Now take a bijection T of X onto itself that preserves mu:
mu(T^-1(A)) = mu(A) for every measurable A. That is, T is a
MEASURE-PRESERVING TRANSFORMATION (abbreviated mpt). It is reversible (T has an
inverse), it is time, and mu is a stationary distribution.
Example. Let X be a coin and mu be the fair coin measure. Then T is the flip
map that switches heads to tails and back. Trivial but valid.
Example. Let X be the unit interval [0,1) with Lebesgue length as mu. Let T(x)
be the fractional part of 10x, i.e. T(x) = 10x - floor(10x). This is the
"decimal shift" map. It preserves Lebesgue length. It is invertible if we
work with two-sided decimal expansions.
Every statistical-mechanical system with a stationary distribution is of this
form: X = set of states, T = one step of time, mu = the equilibrium measure.
The SMOOTH REALIZATION PROBLEM asks:
Which mpts can be drawn using ordinary geometry?
"Ordinary geometry" means: X is (or is measurably bijective to) a compact
smooth manifold M of finite dimension d; mu is (or is pushforward to) a smooth
positive probability volume; T corresponds to a C^infinity diffeomorphism S
of M preserving that volume.
DIFFEOMORPHISM: a bijection M -> M that is C^infinity and whose inverse is
also C^infinity. So it is smooth, invertible, and its inverse is smooth too.
DIRECTIONS ON A MANIFOLD: at every point x of M, the space of directions you
can travel is a d-dimensional real vector space called the tangent space at x,
written T_x M. The derivative of a diffeomorphism S at x is a linear map
D S_x : T_x M -> T_{S x} M. This is the linear approximation of S near x.
SMOOTH POSITIVE DENSITY (of a probability measure on M): we say nu is smooth
positive volume if there is a C^infinity function f: M -> (0, infinity) with
integral 1 over M relative to the ordinary volume form |dx_1 dx_2 ... dx_d| in
any chart, and
d nu = f(x) |dx_1 dx_2 ... dx_d|.
Equivalently, nu has density f which is everywhere positive and smooth.
The function f is called the DENSITY of nu. Ordinary volume itself is the
special case f == 1.
Note: the manifold may have a smooth boundary, and no orientation is needed for
anything here. Boundary and nonorientable cases are handled inline.
A STANDARD NONATOMIC PROBABILITY SPACE is any probability space that is
isomorphic (up to null sets) to the unit interval with Lebesgue measure. A
"nonatomic" measure has no atoms: no single point carries positive mass. The
unit interval with Lebesgue measure is the model example, and measurable
isomorphism theory says any two nonatomic standard spaces are the same up to
null sets. So when we say "X is a standard nonatomic probability space" we
mean: for the purposes of this problem, X looks like Lebesgue measure on [0,1].
WHAT ISOMORPHISM MEANS FOR SYSTEMS. Two mpts (X, mu, T) and (Y, nu, S) are
ISOMORPHIC (or measurably conjugate) if there exist invariant conull measurable
subsets X0 of X and Y0 of Y, and a measurable bijection F : X0 -> Y0 with
measurable inverse such that
F o T = S o F (as functions on X0),
and F carries mu (restricted to X0) to nu (restricted to Y0). No continuity of
F is required. Conjugacy is the notion of sameness for mpts.
So the question is: does every mpt (X, mu, T) of a standard nonatomic space
have a smooth positive-volume model in the sense just defined? The paper
answers no.
===============================================================================
2. ENTROPY, IN ONE PAGE
===============================================================================
Suppose you observe the system and ask "which bin is it in?" from a finite
partition P of X into r pieces P_1, ..., P_r. The answer is a random variable
with r possible values, probabilities p_1, ..., p_r. The SHANNON ENTROPY of
that answer is
H(P) = - sum_i p_i log_2 p_i (with 0 log_2 0 := 0).
H(P) is measured in bits. It is largest when all p_i are equal and equal to
log_2 r. It is 0 when P is degenerate (one atom has probability 1).
Now consider a second observation. Partition X by time-1 atoms T^-1(P). That
says "which atom did the system come from at time -1?" Together, P and T^-1(P)
have a joint answer described by a joint distribution on r x r outcomes.
Iterate: at times 0, 1, ..., n-1, you see an r-symbol pattern. The empirical
laws of these n-step patterns, as n grows, generate entropy at some rate. That
rate is
h(T, P) = lim_{n -> infinity} (1/n) H( P v T^-1 P v ... v T^{-(n-1)} P ),
where "v" is join (the coarsest partition refining both). The limit exists by
subadditivity: the sequence H_n = H(P v ... v T^{-(n-1)}P) satisfies
H_{n+m} <= H_n + H_m, and Fekete's lemma gives existence of the limit as a
real number in [0, log r].
Then the KOLMOGOROV-SINAI ENTROPY OF T is
h(T) = sup_P h(T, P),
the supremum over all finite measurable partitions. It is finite when any
finite-entropy partition generates, and it is a conjugacy invariant: two
isomorphic systems have the same h.
INTUITION. h(T) is the number of bits per time step you gain by watching the
system at arbitrarily fine detail. An identically zero map has h = 0. An
independent coin flip at each time has h = 1 bit per step. A rotation is
zero-entropy: on any fine enough grid, the pattern of atoms you see at times
0, ..., n-1 only grows polynomially in n, so per-step entropy goes to zero.
A Bernoulli shift on two symbols (each step is an independent fair coin) has
h = 1.
ENTROPY OF A FINITE RANDOM VARIABLE. If Y is a random variable taking finitely
many values y_1, ..., y_r with probabilities p_1, ..., p_r, then H(Y) is
H of its law. So H(Y) = - sum p_i log_2 p_i. Same notation.
CONDITIONAL ENTROPY. If Y and Z are finite random variables,
H(Y | Z) = sum_{z : Pr(Z = z) > 0} Pr(Z = z) H( law of Y | Z = z ).
This is "entropy of Y once you already know Z." If Y is independent of Z,
H(Y | Z) = H(Y). If Y is determined by Z, H(Y | Z) = 0. Always
H(Y, Z) = H(Z) + H(Y | Z) and H(Y | Z) <= H(Y).
WHY THIS MATTERS FOR SMOOTH MODELS. Ruelle's entropy inequality. If S : M -> M
is a C^1 diffeomorphism of a compact d-dimensional manifold, if the derivative
DS is bounded in norm, and if nu is any S-invariant probability on M, then
h_nu(S) <= sum of positive Lyapunov exponents of S with respect to nu,
and the right side is finite. Ruelle 1978. So any smooth model has h finite.
This is the FIRST restriction on which systems can have smooth positive-volume
models: h must be finite.
A paper of Katok and Thouvenot, and work following it, showed finite entropy
is not sufficient in a certain sense. This paper shows finite entropy is not
sufficient even when the density is smooth and positive: not every
finite-entropy system has such a model.
Zero entropy is the extreme case of finite entropy. The example in this paper
has zero entropy. So even the easiest systems, ones that produce almost no
information per step, can fail to have smooth positive-volume models.
===============================================================================
3. WHAT THE MAIN THEOREM SAYS
===============================================================================
THEOREM (main theorem of this paper).
There is an ergodic invertible mpt (X, mu, T) of a standard nonatomic probability
space such that:
(i) h_mu(T) = 0.
(ii) There is NO measurably-conjugate system of the following form:
M a compact finite-dimensional C^infinity manifold (possibly with
smooth boundary, possibly nonorientable), nu a smooth positive
probability density on M, S a C^infinity diffeomorphism of M preserving
nu.
In words: we produce a single zero-entropy ergodic system. No matter what
dimension you try, no matter what compact manifold, no matter what smooth
positive density on it, no matter what smooth diffeomorphism preserving that
density, you can never make our system isomorphic to it.
The dimension and manifold are not chosen in advance. They are also not chosen
for each counterexample after the fact: the SYSTEM (X, mu, T) is fixed once
and for all, then a model of ANY dimension is ruled out. This is important:
a weaker statement would say for each dimension d there is a system that has no
model of dimension d. That would be easier. The paper proves a single system
has no model in any dimension.
The manifold being allowed to have boundary and to be nonorientable makes the
counterexample stronger, because it closes more doors.
"ERGODIC." An invariant set of the mpt is a measurable A with T^-1(A) = A up
to null sets. Trivial examples are null sets and conulls. A system is ergodic
if all invariant sets are trivial. Equivalently, if mu is an extreme point of
the convex set of invariant probabilities. In our example ergodicity matters
because the zero-entropy argument for the whole system needs a single block to
behave statistically, which ergodicity guarantees.
"ATOMLESS" / "STANDARD." As above: null set-isomorphic to [0,1] with Lebesgue
measure. Every complete separable metric space with a nonatomic probability is
standard; it is the umbrella property under which ergodic theory
"forgets fidgety set-theory decoration."
END OF SECTIONS 0 THROUGH 3.
Next: what was already known, why smooth positive density is harder than
full support, and then the geometric "switch" that powers the proof.
===============================================================================
4. WHAT WAS KNOWN BEFORE THIS PAPER
===============================================================================
Four prior theorems about realization of mpts. They are the ancestors of this
work and explain why the new counterexample is not just a small improvement.
(A) Lind-Thouvenot 1977. Every ergodic invertible mpt of a standard nonatomic
probability space with finite entropy is isomorphic to a Lebesgue-measure-
preserving HOMEOMORPHISM of the two-torus. Homeomorphism means continuous and
bijective with continuous inverse -- not differentiable, not C^1.
What this says: you CAN always realize every finite-entropy system on a compact
manifold, if you allow the transformation to be merely continuous and if you
allow the invariant measure to be ordinary Lebesgue-measure-density-1 on the
torus. Lebesgue on the torus IS a smooth positive density, so the density part
is fine. The gap is the smoothness of the map.
(B) Quas-Soo 2016, extending Lind-Thouvenot. If A is a hyperbolic toral
automorphism whose characteristic polynomial has no root of unity as root, then
A (or any sufficiently high iterate A^N) represents EVERY ergodic mpt of finite
entropy, with measure entropy strictly smaller than the topological entropy of
A, using an A-invariant BOREL PROBABILITY OF FULL SUPPORT on the torus. Full
support means every open set has positive measure. But the invariant Borel
probability can be SINGULAR with respect to Lebesgue: it can live on a set of
Lebesgue measure zero. A singular measure is not a smooth positive density.
(C) Katok-Thouvenot 1997. They constructed ergodic Z^2 actions which have no
LIPSCHITZ realization on a compact space of finite box dimension, even when the
model may carry any invariant Borel probability. This is about ACTIONS OF Z^2
(two commuting generators realized simultaneously), not about a single
transformation. Their obstruction does not answer the problem for one
transformation.
(D) Foreman-Weiss 2022, Kunde 2024, Gerber-Kunde 2025. These are
ANTI-CLASSIFICATION results: the relation "is measurably isomorphic to" is
not Borel on certain classes of smooth transformations. Foreman-Weiss 2022
uses an invariant measure merely EQUIVALENT to volume, not necessarily with
smooth positive density with smooth positive function. These results show that
classifying smooth systems is hard, but they do NOT say that a specific
abstract zero-entropy system lacks a smooth model. Two different questions.
So the paper's contribution is:
The first example of a single ergodic zero-entropy system with NO smooth
positive-DENSITY model on any compact finite-dimensional manifold, where the
density is smooth and strictly positive and the map is C^infinity.
Every one of the four "closest" results above leaves a gap that this paper
closes. Lind-Thouvenot gives homeomorphisms. Quas-Soo gives singular measures.
Katok-Thouvenot gives actions of Z^2. Foreman-Weiss and Kunde give
anti-classification results inside families of smooth systems.
===============================================================================
5. WHY POSITIVE DENSITY IS HARDER THAN FULL SUPPORT
===============================================================================
A full-support measure can be extremely singular: it can be a single orbit's
empirical average spread over a fractal set of measure zero (respecting full
support topologically). Think of the map: pick one point x, look at its
empirical measure along the orbit, and it can be weird. A full-support
invariant measure on the torus can easily be singular with respect to Lebesgue.
A smooth positive density is different in a very strong way: it is
ABSOLUTELY CONTINUOUS with respect to ordinary volume, with density bounded
above and below by positive finite constants on any compact submanifold. It has
no fractal support. It is the closest thing to "uniform spread out" you can
have.
Here is why this matters for the switch construction: at some point in the
proof, we compute a JACOBIAN. A Jacobian is the local volume multiplier of a
smooth map. "1 + o(1) Jacobian" means the map multiplies volumes by 1 plus a
vanishing error. To turn "the map multiplies volumes by 1+o(1)" into "the
output measure is at most 2 times the input measure" you need the input measure
to have a density that is smooth and positive, so that comparing volumes at two
nearby points is controlled. If the measure were singular, a map could have
Jacobian 1 and still concentrate measure by arbitrary factors. If the density
were merely L^1 or continuous, the Jacobian argument would be less robust.
Two consequences in the paper:
(i) The switch inequality (z_k)_*(rho_k on G_k) <= 2 nu is only asserted
for smooth positive densities.
(ii) The theorem does NOT claim that a full-support singular invariant
measure is impossible. In fact Quas-Soo shows the opposite.
So the paper is specifically about the smooth positive density model. Its
counterexample is only consistent with the prior positive results because
each of them relaxes one of the smoothness, positivity, or single-
transformation restrictions.
===============================================================================
6. THE MAIN GEOMETRIC ENGINE: THE SWITCH
===============================================================================
Now we get to the heart of the proof. This is the one construction that the
combination of smoothness and positive density buys you, and it is what the
paper's zero-entropy example is designed to conflict with.
SETUP: we have a compact d-manifold M, a smooth positive probability volume
nu (density f > 0, smooth), a C^infinity diffeomorphism S preserving nu. We
consider times n = 2^k, dyadic. At each scale k we want to construct a certain
"switch map" z_k.
STEP 1: PARTITION INTO CELLS.
Take a partition of M into measurable cells, at most
exp(C B_k n) of them. Here C is a fixed constant depending only on the
manifold, dimension, and a fixed tolerance. The sequence B_k has two key
properties:
- sum_k B_k^16 < infinity (this will let us defeat any symbolic system
that only offers tests at widely-spaced scales);
- at each dyadic scale k, B_k 2^k is large compared to log(2^k) = k
(this lets the derivative estimates work at that scale).
The specific recipe in the paper is B_k = b_k^{1/8}, where b_k comes from
the stationary estimate of section 7 and satisfies b_k >= 2^{-k/10} =
n^{-1/10} and sum_k b_k^2 < infinity. Then B_k >= n^{-1/80}, so
B_k n >= n^{79/80}, and sum_k B_k^16 = sum_k b_k^2 < infinity. So the
cell count exp(C B_k n) is huge but subexponential in n scaled with
B_k -> 0.
STEP 2: SAMPLE A PAIR OF POINTS IN A CELL.
Build the distribution rho_k on M x M as follows. First pick a cell Q with
probability nu(Q). Then pick two independent points x, y uniformly (by nu
conditioned on Q) inside that cell. Equivalently:
rho_k = sum_{Q} ( (nu restricted to Q) x (nu restricted to Q) ) / nu(Q).
This is a probability on M x M. Both marginals are exactly nu.
STEP 3: THE SWITCH MAP z.
On a large measurable subset G_k of M x M (with rho_k(G_k) >= 1 - eta, eta
being a fixed tolerance like 0.01), there is a measurable map
z : G_k -> M such that:
(a) SHADOWING: For (x,y) in G_k, the orbit of z for times 0,1,...,n
stays close to the orbit of x, and for times -n,...,-1,0 the orbit of z
stays close to the orbit of y, with the closeness tending to zero
uniformly as k -> infinity. So z "follows x into the future and follows
y into the past."
(b) VOLUME CONTROL: The pushforward measure
(z)_* (rho_k restricted to G_k, WITHOUT renormalizing) satisfies
(z)_* (rho_k | G_k) <= 2 nu.
Condition (b) is the important one. It says: even though we are picking pairs
(x,y) in cells and applying a map, the image measure is comparable to the
original nu, with factor 2. No concentration.
Contrast with the trivial attempt "z := x" for the future and "z := y" for the
past; that would not be a single point. The math is to construct a single point
z whose orbit shadows x in the future and shadows y in the past. And the
volume control condition says "the solution set of that boundary-value problem
does not shrink significantly."
WHY DOES THIS EXIST AT ALL FOR A SMOOTH VOLUME-PRESERVING MAP?
Because S is hyperbolic ON MOST OF PHASE SPACE (in the Pesin sense, after a
measurable change of coordinate). At almost every point, the tangent space
splits into a contracting direction (the "slow" or "past" space, along which
past iterations of S push vectors apart) and an expanding direction (the
"fast" or "future" space, along which future iterations of S push vectors
apart). The dimension of these spaces and the local velocity at which they
contract or expand may vary from point to point and there may be slow (zero
or nearly zero) directions, but the finite-time analysis handles this: at each
dyadic scale we control the variation using a "triangular cocycle" of the
derivative and a stationary estimate.
Given x and y in the same small cell (so x and y are very close!), and their
future and past respectively being hyperbolic, a Z-point is found by solving a
boundary-value problem: find z whose future "expanding component" agrees with
x's future expanding component, and whose past "contracting component" agrees
with y's past contracting component. The freedom left over (one component at
each end) is used to solve the linear equation. Then you perturb to a nonlinear
solution by a fixed-point argument.
WHY THE JACOBIAN COMES OUT 1 + o(1).
There are two "obvious" reasons the Jacobian might be bad:
- the expanding directions at time n could have huge or tiny volume ratio;
- the contracting directions at time -n could have huge or tiny volume ratio.
But we take TWO complementary switches. z uses the future of x and the past of
y. Also define w using the future of y and the past of x. Then in the pair map
(x, y) |-> (z, w) the large volume ratios cancel EXACTLY. The full derivative
of the pair map has determinant 1 + o(1). Since we take a PAIR of points, the
Jacobian 1+o(1) is exact to leading order.
WHY THE INJECTIVITY MATTERS.
The switch has to be a well-defined map, not a correspondence. Two things are
needed:
- the boundary-value problem has a UNIQUE solution in a large enough
neighbourhood (contraction mapping, uniqueness ball), and
- different inputs (x,y) do not lead to the same output (z,w).
The paper proves both. The uniqueness radius is bigger than the shadowing
radius, and there is a positive wall margin in the cells so outputs come back
into the same cell, and their endpoint bins are recovered from their
endpoint values.
WHAT COULD NOT BE DONE WITHOUT SMOOTH POSITIVE DENSITY.
The injectivity argument uses the density: it needs f > 0 everywhere so that
"the output density at z is at most 2 times the input density at (x,y)". A
measure that is only singular or only absolutely continuous but not smooth
would let the pair map concentrate.
===============================================================================
7. STATIONARY ESTIMATES: SUMMABLE DYADIC ERRORS
===============================================================================
The switch needs, for each dyadic n = 2^k, a small exceptional set of points
where the derivative control of the orbit fails. The paper shows that the
failure set shrinks fast enough that summing over k converges.
The key technical tool is a stationary estimate which we now explain.
A STATIONARY PROCESS. A sequence of real random variables (Y_t)_{t in Z} is
stationary if translating the indices does not change the joint law of any
finite list:
(Y_{t_1 + s}, ..., Y_{t_m + s}) has the same law as (Y_{t_1}, ..., Y_{t_m}).
So it is invariant in law under the shift map. Examples: i.i.d. sequences,
periodic sequences with random phase, orbit of a smooth function along a
measure-preserving T.
BLOCK AVERAGES AND CHORDS. For h >= 0, let
A_h = (1/2^h) sum_{t=0}^{2^h - 1} Y_t
be the average on the first 2^h terms, and let Z_h be
Z_h = (1/2^h) max_{0 <= u <= 2^h} | sum_{t=0}^{u-1} Y_t - u A_h |.
Z_h measures how much the cumulative sum drifts away from the straight chord
of slope A_h. This is exactly the deviation of a random walk from its chord,
at scale 2^h.
LEMMA (dyadic chord estimate). For a real stationary process with finite second
moment, define
v_l = E(A_{l-1}^2) - E(A_l^2), l >= 1.
Then v_l >= 0, and sum_{l >= 1} v_l <= E(Y_0^2), and for h >= 1,
||Z_h||_{L^2} <= (1/2) sum_{a=0}^{h-1} 2^{-a/2} sqrt(v_{h-a}).
Consequence: sum_{h >= 1} ||Z_h||_{L^2}^2 <= C E(Y_0^2) for an absolute
constant C.
IMPORTANT: this is a SUM OF SQUARES being finite, not just a sup. That is what
gives summable dyadic control: at scale k, the L^2 error in the "chord
deviation" is bounded by a quantity whose SQUARED sum is finite.
INTERPRETATION. Think of the random walk S_n = Y_0 + ... + Y_{n-1} drifting at
speed A_h at scale 2^h. The lemma says the difference between S_n at the
intermediate time and the linear interpolation of S_0 to S_{2^h} has L^2 norm
summable in h. That is a strong statement: it is the smooth-dynamics
equivalent of "a martingale has a quadratic-variation-driven central limit",
but in L^2 rather than distributional form.
THE APPLICATION: TRIANGULAR COCYCLE.
On a d-manifold with an S-invariant probability, consider the orthonormal
FRAME BUNDLE F M. That is: over each point x is a fiber of orthonormal
identifications e : R^d -> T_x M. This is a finite-dimensional bundle. For
any frame e over x, form the positive-diagonal QR factorization
D S_x o e = tilde S(e) o R(e),
where tilde S(e) is a new frame over S x and R(e) is upper triangular with
positive diagonal. This is ordinary Gram-Schmidt: R is the triangular shape of
derivative, tilde S(e) is the orthonormalized result. The map e |-> tilde S(e)
is a homeomorphism (indeed C^infinity) of the frame bundle.
Then, choosing a tilde S-invariant probability tilde nu on the bundle that
projects down to the original nu, the quantities g_j(e) = ln(R(e)_{jj}) are
bounded continuous functions (compactness of F M gives uniform boundedness of
R(e) and R(e)^{-1}) and
(g_j(tilde S^t e))_{t in Z}
is a real stationary process.
Applying the dyadic chord lemma to each g_j and summing gives, for each
dyadic scale k and each tolerance eta, a compact "good set" of base points in
M (call it G_k) with nu(G_k) >= 1 - eta/100, such that on G_k the cumulative
log-diagonals of the triangular cocycle deviate from their chord at most b_k n,
where b_k is the small positive sequence b_k >= 2^{-k/10} with
sum_k b_k^2 < infinity. This is the same b_k whose eighth root B_k
controls the cell count in section 6.
This gives the "good finite orbit segments" property: within the good set,
derivative growth along the orbit is uniform on the fast directions and on
the slow directions, at every subinterval of a good segment, with a small
additive error that is summable over dyadic scales.
WHY THE SUMMABLE-SQUARED RATE MATTERS LATER.
Thematic reason: at each dyadic scale k you can afford an exceptional set of
mass O(b_k^2). Since sum_k b_k^2 < infinity, the "good points at all large
enough scales" still have full measure. If the rate were not summable, the
"good at all scales" set could have zero measure. This is exactly what a
pointwise ergodic theorem does NOT give you: an ergodic theorem gives
convergence almost surely along a sequence, but the RATE of convergence can be
arbitrarily slow. The dyadic chord lemma gives an L^2 rate.
===============================================================================
8. LINEAR ESTIMATES: HOW TO SPLIT THE TANGENT SPACE
===============================================================================
Given a good dyadic segment, we know the diagonal triangles of the derivative
are close to a chord. But we need more: we need to split the d-dimensional
tangent space into a "fast" part (expanded by the future) and a "slow" part
(expanded by the past), with two properties:
- the split varies continuously with the point on the segment, and
- the split has controlled angle (the projections between the two parts
are bounded in norm).
The paper constructs this split in three steps.
STEP 1: SPLIT THE DIAGONAL RATES.
The diagonal entries of the triangular cocycle are e^{lambda_j} at each time
step. Order them so that lambda_{(1)} >= lambda_{(2)} >= ... >= lambda_{(d)}.
Choose any real threshold gamma such that no lambda_j is within distance g of
gamma (g > 0 separation). Let F be the set of indices j with lambda_j >= gamma
+ g ("fast"). Let p = |F|, so p is the dimension of the fast part.
Define:
Lambda = sum_{j in F} lambda_j (fast rate sum)
Omega = - sum_{j not in F} lambda_j (slow rate sum, note negative)
STEP 2: CONJUGATE TO HAVE INVARIANT BLOCKS.
The triangular cocycles are triangular, but they mix coordinates. The paper
conjugates A_t -> B_t = C_{t+1}^{-1} A_t C_t where C_t are upper-triangular
changes of basis, chosen so that A_t becomes block-diagonal with the two
groups F and complement F. This is done by processing super-diagonals in
increasing order and using elementary "shear" matrices Q_t = Id + u_t E_{ij}
at each off-block position. Each shear is solved via a one-step linear
recurrence u_{t+1} = q u_t + f_t with |q| < 1 or |q| > 1 (the separation of
rates guarantees |q| != 1). The C_t are uniformly bounded by e^{C B n}, since
the shear bounds are polynomial in n and ln(n) <= B n.
STEP 3: ESTIMATE EXTERIOR POWERS.
After conjugation, the derivative splits into a fast block (upper triangular,
dimension p) and a slow block (upper triangular, dimension d-p). Then:
- the fast block's p-form growth rate is Lambda l + O(B n);
- the slow block's (d-p)-form growth rate is Omega l + O(B n);
- mixed forms with at least one slow vector in a p-form product have growth
rate (Lambda - 2g) l + O(B n);
- and similarly backward with Omega - 2g.
The gap "2g" is the KEY. The mixed forms are strictly slower by 2g than the
pure fast p-form. This makes the leading p-form DOMINATE: the leading minor
is roughly e^{Lambda n}, every mixed minor is roughly e^{(Lambda-2g)n}, so
their ratio is e^{2g n} and the leading term is exponentially bigger than
every other. In the Jacobian computation the leading term therefore
determines the sign and magnitude of the full determinant, and the other
minors are relatively negligible.
The 2g is the separation margin between "fast" and "slow" rates. It is what
allows the split to be meaningful and to produce a well-conditioned Jacobian.
STEP 4: COMMON SPACES IN A SPATIAL BIN.
At scale k, M is partitioned into SPATIAL BINS: small balls (or boxes in
charts) of radius roughly h_0 = e^{-B n}. For each bin U, we need to choose a
SINGLE threshold gamma_U and a SINGLE pair of reference spaces E_U^s, E_U^f
that work for ALL good points in U.
This works because:
- all good points in U have their sorted rate lists agreeing to within
o(g) (via the chord control and the short derivative comparison);
- there are only d+1 candidate thresholds (1+8j)g for j = 0,...,d, and at
most one of them is within 3g of a given rate;
- so there is a candidate that separates every list with margin > 2g.
Given a threshold, the reference spaces are chosen once per bin (via a fixed
good witness in the bin). All actual spaces in the bin are within e^{-c g L}
of the reference, where L is a short dyadic length (roughly n b^{1/4}, larger
than the dominant short error). This is because of a "short map comparison"
argument: at scales short compared to n, the derivative of S^L at nearby
points has nearly identical singular vectors, with error decaying like e^{-Bn}.
STEP 5: RELATIVE STABILITY UNDER PERTURBATION.
The Jacobian argument uses p-form products over LONG times. These products have
ASTONISHINGLY small volumes (e^{Lambda n}, Lambda can be very negative for a
contracting direction). We need: if a nearby point has slightly perturbed
derivative, its p-form product is still comparable to the unperturbed one. This
is a relative perturbation estimate:
|| bigwedge^p Psi(m,0) - bigwedge^p A(m,0) || <= e^{sigma m + a}
* ((1 + C epsilon e^{a+|sigma|})^m - 1),
where the A_t are the true matrix, the Psi_t = A_t + E_t are the perturbed
ones, epsilon is the size of the perturbation, and sigma is the volume
growth rate. This inequality has a factor e^{sigma m + a} AND a factor
(1 + c)^m - 1. The 1 + c is where the perturbation acts. So if m epsilon is
small (perturbation is very small for many steps), the "1 + c" stays close to 1.
This allows the Jacobian argument to be robust: even though the volume in a
given p-form is microscopic, its PERTURBATION is even smaller by a factor
that tends to zero.
NEXT: turn the linear data into the actual geometric construction. See section
9 below.
===============================================================================
9. BUILDING THE SWITCH: CELLS, JOINS, JACOBIAN
===============================================================================
Sections 7 and 8 gave us the linear data at every dyadic scale. Now we turn
that into the actual geometric construction claimed in section 6. The proof is
in five steps.
STEP A: FIX SCALES AND THRESHOLDS.
At dyadic scale k, n = 2^k. Pick the same b_k as before (from the stationary
estimate), and define:
B = b_k^{1/8} (used for cell sizes and counts)
g = b_k^{1/2} (rate separation margin)
L = ceil(b_k^{1/4} n) (short comparison time)
h_0 = e^{-B n} (coarse bin side)
r = e^{-2 B n} (product cell side, much smaller than h_0)
delta= e^{-3 B n / 2} (contraction uniqueness radius)
The scale hierarchy is:
log n < b_k n < g L < g n < L < B n,
with each "<" meaning o( ), and the gap growing.
Also fix a tolerance eta (like 0.01). And a small number tau (like 0.001) for
wall margins inside cells. Neither depends on k.
STEP B: THREE PARTITIONS.
At scale k we work with three partitions of M.
1. COARSE BINS. Cover M with finitely many coordinate charts, each with a
smaller "inner box". Inside each inner box, grid into coordinate cubes
of side h_0. These are the coarse bins. Their total number is at most
C h_0^{-d} = exp(C d B n) = exp(O(B n)).
2. PRODUCT CELLS. For each coarse bin Q that meets the good set at scale k,
we have a plain threshold gamma_Q and reference spaces E_Q^s, E_Q^f
(section 8, STEP 4). In the coordinate chart of Q, tile by
parallelepipeds spanned by these two spaces with side length r. These
are the product cells. The forward expansion of the change-of-basis map
is at most sqrt(2) (E_Q has orthonormal columns), so diameter is at most
C_d r. The inverse norm is at most e^{C B n}, so the volume of one
product cell is at least e^{-C B n} r^d up to constants. Total number of
product cells is at most exp(C B n) r^{-d} = exp(C' B n).
3. ADJUSTED CELLS. Add one extra cell for all omitted points and bins not
meeting the good set. This is the partition D_k in section 6.
EFFECTIVE PARTITION SIZE. #D_k <= exp(C B_k n).
STEP C: WHICH PAIRS SURVIVE.
A pair (x, y) in M x M is "retained" if it lies in the square of one product
cell Q' x Q' and in addition:
- both x and y lie in the good set at scale k;
- both x and y are in the interior of their product cell, at distance at
least tau r from each of the product-coordinate faces;
- S^n x, S^{-n} x, S^n y, S^{-n} y each lie at distance at least tau h_0
from the faces of their coarse endpoint bins;
- if M has boundary, no iterate at integer time in [-n, n] is within
c_0 h_0 of the boundary, for a fixed c_0.
The volume of points excluded by wall strips is at most C_d tau of nu (times
fixed constants), and the volume of points excluded by any other condition is
at most eta/100 of nu, uniformly. So rho_k(G_k) >= 1 - eta for a fixed eta:
removing wall strips and neighborhoods of the boundary and bad points costs
at most eta in mass.
STEP D: THE BOUNDARY-VALUE PROBLEM.
For a retained pair (x, y):
- Define four endpoint projections:
F_x(u) = P^+_{Q,U_x} o chi_{U_x} o S^n(u)
F_y(u) = P^+_{Q,U_y} o chi_{U_y} o S^n(u)
P_x(u) = P^-_{Q,V_x} o chi_{V_x} o S^{-n}(u)
P_y(u) = P^-_{Q,V_y} o chi_{V_y} o S^{-n}(u)
Here chi_U is the fixed localization map in the coarse bin U; P^+ is the
orthogonal projection onto the fast space of the reference endpoint bin;
P^- is the orthogonal projection onto the slow space of the past endpoint
bin.
- Seek z and w satisfying:
F_x(z) = F_x(x), P_y(z) = P_y(y)
F_y(w) = F_y(y), P_x(w) = P_x(x).
That is: z's future "arrives where x's future arrives" (from the fast
perspective) and z's past "departs from where y's past departs" (from the
slow perspective). Similarly for w with x and y exchanged.
This is eight equations in the joint unknowns z and w. But z alone solves a
closed system (first line), and w alone solves a closed system (second line).
Each is 2n time steps long. The linearized version (around the reference
orbits of x and y) has a matrix form.
STEP E: THE LINEAR INVERSE.
Write the linearized system in "weighted" coordinates. On the future half, put
v_tilde_t = e^{-gamma t} v_t. This strips out the average growth. Then:
- on the slow direction (expanded by past iterations of S), the weighted
future propagator has norm at most e^{C B n} e^{-g |t|}. It decays as we
move forward in time;
- on the fast direction (expanded by future iterations of S), the weighted
past propagator decays as we move backward in time.
So the future side propagates "well" in one direction, the past side in the
other. Composing, all forced terms are bounded by e^{C B n} times the sup-norm
of the data, and the two "kernel graph" corrections have norm at most
e^{-2 g n}.
Then the map (matching at time 0) is an isomorphism from
E_0^s(x) oplus E_0^f(y) onto the tangent space at the starting point. The
inverse has unweighted norm at most Q_k = e^{C g n}. That is the crucial
bound: Q_k is exponential in g n but g n is much smaller than B n (by the
scale hierarchy), so Q_k r and Q_k delta both tend to zero. This is what
makes the contraction argument work.
STEP F: NONLINEAR SOLUTION BY CONTRACTION.
The nonlinear problem is v = L^{-1}(a_0 + R(v)), where a_0 is the inhomogeneous
data of size O(r) and R(v) is the quadratic Taylor remainder of the step maps.
Because all two-variable second derivative bounds are uniform (compactness of
M and its charts), there is a fixed constant C with
|R(v)| <= C |v|^2 and
|R(v) - R(u)| <= C delta |v - u| for |v|,|u| <= delta.
The map v |-> L^{-1}(a_0 + R(v)) sends the delta-ball into the delta/2 ball
and is a contraction. So there is a unique fixed point v* in the delta-ball,
with |v*| <= 2 Q_k r. Its base point is z. Because Q_k r << delta and delta
<< h_0, the actual orbit of z stays in M and in a fixed neighbourhood of the
reference orbits. The shadows are close to x in the future and y in the past,
with error at most e^{C g n} r, which tends to 0.
In product coordinates, z differs from the "ideal exchange" z_0 = (y_s, x_f)
by at most o(r). Same for w. And z, w both stay in the original product cell,
at distance at least 3 tau r / 4 from each face, and their endpoints stay in
the four original coarse endpoint bins.
STEP G: THE JACOBIAN.
Now the key: compute the Jacobian of the full pair map (x, y) |-> (z, w).
Define H_{ab}(u) := (F_a(u), P_b(u)) for a, b in {x, y}. This is a
d-dimensional map from the manifold to a product of two Euclidean spaces of
dimensions p_Q and d - p_Q.
Claim: the four derivatives DH_{xx}(x), DH_{yy}(y), DH_{xy}(z), DH_{yx}(w) are
all invertible, and their determinants satisfy (up to 1 + o(1)):
|det DH_{xx}(x)| = A_x B_x
|det DH_{yy}(y)| = A_y B_y
|det DH_{xy}(z)| = A_x B_y
|det DH_{yx}(w)| = A_y B_x
where A_x = leading p-form minor of DS^n at x, B_x = leading (d-p)-form minor
of DS^{-n} at x, etc. The cross-terms cancel because the endpoint projections
were chosen to isolate one form at each end.
Then the exact identity:
|det D(z,w)(x,y)| = |det DH_{xx}(x)| |det DH_{yy}(y)|
/ |det DH_{xy}(z)| |det DH_{yx}(w)|,
which is exactly the ratio 1 + o(1): A_x B_x A_y B_y / (A_x B_y A_y B_x) = 1.
The relative perturbation lemma (section 8, STEP 5) tells us the perturbations
of the four minors do not destroy this: |det DH_{xy}(z) - A_x B_y| / (A_x B_y)
-> 0, similarly for the others. So the Jacobian is genuinely 1 + o(1), not just
"approximately one in average".
STEP H: PUTTING IT TOGETHER: MEASURE INEQUALITY.
Take the countable measurable partition of G_k into pieces where the cell and
the four endpoint buckets are fixed. On each piece, (x, y) |-> (z, w) is a
smooth local diffeomorphism with Jacobian 1 + o(1).
Apply the change-of-variables formula, and sum over pieces. Because injectivity
holds (see below), the images are disjoint. And the product of two smooth
positive densities has bounded ratio to any nearby density, so summing gives:
(z_k)_* (rho_k on G_k) <= 2 nu.
This is (b) of the switch theorem.
STEP I: INJECTIVITY.
Suppose (z, w) has two preimages (x, y) and (x', y'). First, cell Q must be the
same because both outputs are in the interior of their common product cell and
cells have disjoint interiors. That recovers Q and the choice of reference
spaces E_Q^s, E_Q^f.
Then the four endpoint bins U_x, V_x, U_y, V_y are recovered from the actual
endpoint values: bin(S^n z), bin(S^{-n} w), bin(S^n w), bin(S^{-n} z). Positive
wall margins ensure this recovery is stable.
So the competitor (x', y') must have the same cell, same reference spaces, and
same four endpoint projections.
Then x and x' both lie near z's future and near w's past; y and y' both lie
near w's future and near z's past. So all four points are close. But they must
satisfy the same boundary equations, so a direct application of the
uniqueness-in-delta-ball argument (from STEP F) with matching datum 0 and
terminal data 0 forces x = x' and y = y'.
The uniqueness uses the delta-ball, which is much bigger than the shadowing
radius, so this is robust.
STEP J: MEASURABILITY.
The retained set G_k is Borel. The switch (x, y) |-> (z, w) is measurable
because it is constructed as the fixed point of a continuous map parametrized
measurably by (x, y) (on countably many Borel "locally uniform" pieces). The
Jacobian identity gives the "restriction of a smooth local diffeomorphism"
property on each piece.
===============================================================================
NEXT: the SYMBOLIC side. We build a zero-entropy system with large window
entropy and a graph relation between windows. Then we compare the two sides.
===============================================================================
10. THE SYMBOLIC SIDE: FOUR WORD PROPERTIES
===============================================================================
Now we change subjects completely. We build a symbolic (finite-alphabet) system
that is going to be the zero-entropy counterexample. It will have two crucial
properties:
- its Kolmogorov-Sinai entropy is zero;
- short lists of its legal "words" have very large Shannon entropy.
And it will have a specific algebraic relation between words at past and future
windows, which the smooth switch cannot satisfy.
This section constructs the words. The next section builds the system.
---------------------------------------------------------------------------
THE ALPHABET: NO, WE NEED VECTOR LABELS.
---------------------------------------------------------------------------
Let s be a positive integer. Let F_2 = {0, 1} with addition mod 2. Let F_2^{2s}
be the (2s)-dimensional vector space over F_2. On F_2^{2s} put a nondegenerate
ALTERNATING FORM (a "symplectic pairing"):
[u, v]_s = sum_{a=1}^{s} ( u_a v_{s+a} + u_{s+a} v_a ) in F_2.
This is a binary-valued function: it takes each pair of vectors to 0 or 1. It
is linear in each argument. It is alternating: [u, u]_s = 0 always.
Nondegeneracy: if [u, v]_s = 0 for every v then u = 0.
Pick s with K >= 256 where K = floor(sqrt(ell)) for some ell defined below.
Take the alphabet A := A_1 := F_2^{2s_1} \ {0}, the nonzero vectors of F_2^{2s_1}.
Each letter of A is a 2s_1-bit string. And a word is a finite sequence of such
strings.
---------------------------------------------------------------------------
THE HIERARCHY: LEVELS OF WORDS.
---------------------------------------------------------------------------
We build a sequence of alphabets and word families:
level 1: alphabet A_1 (nonzero vectors of F_2^{2s_1}), each letter has length
2s_1 bits. There are L_1 = 2^{2s_1} - 1 letters. Each is called a
level-1 word. m_1 = 1 (level 1 words are one letter long).
level 2: a family W_2 of L_2 words, each made of N_1 letters from A_1. Each
level-2 word has m_2 = m_1 N_1 = N_1 symbols.
level 3: a family W_3 of L_3 words, each made of N_2 words from W_2. Each
level-3 word has m_3 = m_2 N_2 = N_1 N_2 symbols.
...
level i: a family W_i of L_i words, each made of N_{i-1} words from W_{i-1}.
Each level-i word has m_i = m_1 N_1 N_2 ... N_{i-1} symbols.
The scaling parameters are chosen so that L_i and m_i grow in controlled ways:
eps_i = 2^{-i-2}
L_i = 2^{2s_i} - 1
ell_i = log L_i
K_i = floor(sqrt(ell_i))
N_i = 2^{K_i}
H_i = 2^{floor(K_i/2)}
D_i^* = 2^{floor(K_i/8)}
m_1 = 1, m_{i+1} = m_i N_i
and the (s_i) are recursive: 2s_{i+1} is the greatest even integer at most
X_i := (eps_i / 100) * N_i * ell_i.
We need K_1 >= 256 to get this started. Nothing depends on the exact value of
s_1 beyond that.
Once level i is built, we identify its L_i words with the nonzero vectors of
F_2^{2s_i} via an arbitrary fixed bijection, so each level-i word u carries a
label vector, and we can test [u, v]_i for u, v in W_i.
For a word w in W_{i+1}, write c_i(w) for its list of constituent level-i words:
c_i(w) = (u_0, u_1, ..., u_{N_i-1}) with each u_j in W_i. Unambiguous because
all level-i words have length m_i.
Note: later, N_i plays the role of "how many constituents in a level-(i+1)
word"; H_i is the "group size" used in the graph condition; D_i^* bounds the
window lengths and prescribed power-of-two lags, all MUCH smaller than N_i.
---------------------------------------------------------------------------
THE FOUR PROPERTIES.
---------------------------------------------------------------------------
We want each family W_i of L_i words, each of length m_i, to satisfy:
(W1) GRAPH RELATIONS WITHIN GROUPS.
For every w in W_i (or W_{i+1}, depending on level conventions), with
c_i(w) = (u_0, ..., u_{N_i-1}), and for every 0 <= b < N_i / H_i,
every 0 <= k <= floor(K_i / 8), and every 0 <= r < r + 2^k < H_i,
[ u_{b H_i + r}, u_{b H_i + r + 2^k} ]_i = 0.
That is: within each "group" of H_i consecutive constituents, positions
separated by a power of two must be orthogonal (with respect to the
symplectic form at level i).
(W2) DISTANCE BETWEEN WORDS AND SHIFTED WORDS.
If v, w in W_{i+1} are distinct, then
d_H(c_i(v), c_i(w)) >= (1 - eps_i) N_i
where d_H counts how many constituent positions differ.
More strongly: for all v, w, z in W_{i+1} (repetitions allowed) and every
offset 1 <= t < N_i,
d_H( (c_i(v) c_i(w)) [t, t + N_i), c_i(z) ) >= (1 - eps_i) N_i.
Here (c_i(v) c_i(w)) is the concatenation of the two constituent lists,
and [t, t+N_i) cuts a shifted window of length N_i. So any window crossing
the boundary between two concatenated words looks very different from any
legal word. That is comma-free-ness.
(W3) DISTRIBUTIONS OF SHORT WINDOWS.
Let U be uniform on W_{i+1}. For each 0 <= t <= N_i - D_i^*, there is a
probability Q_{i,t} on W_i^{D_i^*} such that
TV( law of c_i(U)[t, t + D_i^*), Q_{i,t} ) <= 10^{-4}.
Moreover the Q_{i,t} satisfy an atom bound: for 1 <= d' <= D_i^*,
distinct positions 0 <= r_1 < ... < r_{d'} < D_i^*, and any prescribed
labels v_1, ..., v_{d'} in W_i,
Q_{i,t}{ u : u_{r_j} = v_j for all j } <= 2 * (2^{2 K_i} / L_i)^{d'}.
The right side is a small number -- exponentially small in d' * K_i.
(W4) APPROXIMATION IN EACH INDIVIDUAL WORD.
For every fixed w in W_{i+1}, write c_i(w) = (u_0, ..., u_{N_i-1}) and
take J uniform on {0, ..., N_i-1}. Suppose U is uniform on W_i. Then for
each 0 <= t <= N_{i-1} - D_{i-1}^*,
TV( law of c_{i-1}(u_J)[t, t + D_{i-1}^*), law of c_{i-1}(U)[t, t +
D_{i-1}^*) ) <= 10^{-4}.
That is: short internal windows of a fixed parent word look statistically
like short windows of a uniform lower-level word, up to 10^{-4} in
total variation. This is a per-word condition, not an average over all
of W_{i+1}. This is what makes the eventual invariant measure "see"
near-uniform short windows even when it weights some parents more than
others.
Total variation distance: (1/2) * sum over a of |p(a) - q(a)|.
---------------------------------------------------------------------------
THE FOUR PROPERTIES ARE COMPATIBLE: A PROBABILISTIC-EXISTENCE PROOF.
---------------------------------------------------------------------------
Proposition 10.1 (Finite word families, restatement).
For every s_1 with K_1 >= 256, the recursion above produces families W_i of L_i
words, each of length m_i, satisfying (W1)-(W4) at every level.
Sketch of proof, in four stages.
Stage 1: SAMPLE LISTS WITH THE GRAPH RELATIONS.
Fix W_i and its vector labeling. Generate the constituent list
(U_0, ..., U_{N_i-1}) of a W_{i+1} word in independent groups of H_i
consecutive positions. Within each group, process from left to right; choose
each new label UNIFORMLY among the nonzero vectors orthogonal to all earlier
labels at the specified power-of-two lags within the same group. There are at
most K_i linear constraints at any point, so the kernel has codimension at most
K_i, so the number of allowed labels is at least
2^{2s_i} / 2^{K_i} (roughly),
which is a big number since 2s_i is much bigger than K_i.
Call this sampling law P_i^0. Every list it produces satisfies (W1). Also P_i^0
gives each individual position a UNIFORM marginal on W_i: this is because the
symplectic group Sp(2s_i, F_2) acts transitively on nonzero vectors, and the
sampling rule is equivariant under that action.
Stage 2: ENFORCE (W4) BY CONDITIONING.
For each parent word, look at the empirical distribution of short windows from
the constituent labels. If it is too far from the uniform law, toss the parent
out. A lemma says: if Y_1, ..., Y_q are independent random probability vectors
on a set of size A, all with the same mean p, then
E TV( (1/q) sum Y_b, p ) <= (1/2) sqrt(A/q).
This is a Chebyshev-like bound which we apply to the q_i = N_i / H_i independent
groups. Markov + union over the N_{i-1} offsets gives: the probability that
some parent word violates (W4) is less than a quantity F_i that we will verify
is small.
Let E_i be the event that all (W4) requirements hold. Then P_i^0(E_i) >= 1/2
(once we verify F_i < 1/2). Define P_i := P_i^0( . | E_i ).
Stage 3: ENFORCE (W3) AND (W2) BY FRESH SAMPLING.
Sample M_i := L_{i+1} independent lists with law P_i. For each offset, the
empirical law of short windows over the samples estimates Q_{i,t}. Another
application of the averaging lemma + Markov + union over N_i offsets shows that
(W3) fails with probability < 1/4.
For (W2): reference law = each distinct list is uniform on W_i^{N_i}. A "distance
failure" means two lists agree at more than eps_i N_i positions, or a shifted
window agrees with a legal word at more than eps_i N_i positions. For any fixed
set of matching positions, form a multigraph whose vertices are (list-index,
position) pairs and whose edges are the asserted equalities. Each connected
component with v vertices imposes at least v-1 equalities: probability at most
L_i^{-(v-1)}. So a set of e matching positions has probability at most
L_i^{-e/2}. Summing over all possible match sets and comparing to P_i via a
likelihood ratio gives a union bound T_i, which we verify is small.
Stage 4: COMBINE.
The union bound shows: at each level, the probability that a randomly sampled
family fails ANY of (W4), (W3), (W2) is < 1. So the family exists with all
properties. Detail: at most (1/2) + (1/4) + (1/4) = 1 of mass can fail, so
some family works.
Finally: because (W2) with v != w forces distance N_i, the sampled words are
distinct (else distance would be 0), and there are exactly L_{i+1} of them.
That is our family W_{i+1}.
The bounds F_i, E_i, T_i are verified to be < 1/2, 1/4, 1/4 at every level via
explicit numerical inequalities in the paper. They hold because K_i grows
extremely fast (K_{i+1} >= 2^{K_i/3} >= K_i^2 >= 4 K_i).
---------------------------------------------------------------------------
CONSEQUENCE 1: RECOGNITION OF WORDS AND ALIGNMENT.
---------------------------------------------------------------------------
Lemma 10.2 (Recognition). Define
d_i = prod_{j=1}^{i-1} (1 - eps_j),
d_* = prod_{j=1}^{infinity} (1 - eps_j) > 0.
Note d_* >= 1 - sum_j eps_j = 3/4.
(a) Distinct v, w in W_i satisfy d_H(v, w) >= d_i m_i >= d_* m_i.
(b) For all v, w, z in W_i and every offset 1 <= t < m_i,
d_H((vw)[t, t + m_i), z) >= d_i m_i >= d_* m_i.
Proof. The finite products d_i are decreasing and converge to
1 - sum eps_j = 3/4 > 0. The bound on d_H(v,w) follows by induction from
(W2) applied to successive levels: at level i+1, two distinct words differ in
at least (1 - eps_i) N_i constituent positions; each difference contributes at
least d_i m_i symbol mismatches at level i; that gives d_{i+1} m_{i+1}.
The offset case: if m_i divides t, the constituent offset t / m_i is nonzero,
and (W2) applies directly. If not, each aligned target constituent faces a
length-m_i source window at a nonzero offset modulo m_i. That source window
crosses exactly two consecutive legal level-i words, and the induction
hypothesis (part b at level i) gives at least d_i m_i mismatches per target
constituent.
This is the "comma-free index" property: a legal word, shifted by any nonzero
amount and viewed against another legal word, looks different by at least a
fixed fraction of its length. This is what lets a smooth shadowing orbit
recover word labels.
---------------------------------------------------------------------------
CONSEQUENCE 2: THE RATES R_i DECAY, AND A 16-TH POWER DIVERGES.
---------------------------------------------------------------------------
Lemma 10.3 (Rates). Define R_i = ell_i / m_i (entropy per symbol at level i).
Then:
(a) m_i is a power of two, log m_i = sum_{j<i} K_j.
(b) R_{i+1} <= (eps_i / 100) R_i, so R_i -> 0.
(c) R_i >= R_1 * 200^{-(i-1)} * 2^{-(i-1)(i+20)/2}, so R_i decays at most
exponentially times a quadratic-in-i exponential correction.
(d) sum_{i >= 1} K_i R_i^{16} = infinity.
The last statement is the crucial "divergence": K_i R_i^{16} does not decay.
Physically: each level offers K_i "test scales" (dyadic positions inside a
word), and each scale carries entropy R_i per symbol. Sum over all scales is
infinite in the 16th-power sense.
Combining (b) with the growth K_{i+1} >= 2^{K_i/3} >= K_i^2:
for large i, log(K_i R_i^{16}) >= 2^{i-1} log K_1 + 16 log R_1 - 8(i-1)(i+20),
which tends to +infinity.
The comparison between smooth and symbolic is now set up: on the smooth side
the sum of B_k^{16} is FINITE, and on the symbolic side sum K_i R_i^{16} is
INFINITE. This "summable vs. non-summable" is where the eventual contradiction
will live.
===============================================================================
11. THE PROCESS: ZERO ENTROPY, BIG WINDOW ENTROPY
===============================================================================
The word families of section 10 are the ingredients. Now we glue them into a
single ergodic mpt with two crucial properties:
(a) Kolmogorov-Sinai entropy zero;
(b) short window entropy almost maximal.
That is: per-symbol entropy is tiny, but a short window of D consecutive
level-i words carries almost D * ell_i bits of entropy. And (a) and (b) can
both be true because "almost maximal" is offset by "per level i the maximum
per symbol is R_i which tends to zero".
CONSTRUCTION: THE COMPACT PHASE EXTENSION.
Let A = W_1 be the level-1 alphabet (nonzero vectors of F_2^{2s_1}). Take the
compact space
Omega = A^Z x prod_{i >= 1} (Z / m_i Z).
A point is (omega, a) with omega: Z -> A a bi-infinite sequence of letters and
a = (a_1, a_2, ...) with a_i in Z/m_i Z. The coordinate a_i has a MEANING: it
is the position of the origin relative to the boundary of the level-i word
containing it.
Define a subspace X of Omega by the conditions:
(i) COMPATIBILITY. a_{i+1} = a_i mod m_i for every i. So higher phases
refine lower phases.
(ii) LOCAL LEGALITY. Whenever t = -a_i mod m_i (i.e. t is a level-i word
boundary), the length-m_i block
(omega_t, omega_{t+1}, ..., omega_{t+m_i-1}) is a legal level-i word,
i.e. belongs to W_i.
X is closed in Omega (finite constraints on each coordinate from a closed set),
so X is compact. On X the shift
T(omega, a) = (omega', a'), omega'_t = omega_{t+1}, a'_i = a_i + 1 mod m_i
is a homeomorphism.
NONEMPTY. For any j, take a word from W_j and align it bi-infinitely with
boundary at 0. Since higher-level words are concatenations of lower-level
ones, this sequence satisfies all conditions through level j. These finite-
level sets are nested; compactness in X gives an infinite-point intersection,
which is nonempty. Alternatively: by Konig's lemma there is an infinite path
through the tree of consistent finite sequences.
INVARIANT MEASURES EXIST. Weak limits of empirical averages
(1/n) sum_{j=0}^{n-1} delta_{T^j x} are T-invariant, so the compact convex set
M_T(X) of T-invariant probabilities is nonempty. Standard ergodic theory
gives: this convex set has extreme points, and extreme points are exactly the
ergodic measures.
PROPOSITION 11.1 (Invariant system on X).
There is a T-invariant ergodic probability mu on X. Every T-invariant
probability on X is nonatomic, is standard modulo null sets, and has
Kolmogorov-Sinai entropy zero.
Proof pieces.
Ergodicity: pick a countable uniformly dense family of continuous test
functions. Successively restrict to the compact face of invariant probabilities
where each f_j integrates to its maximum over the previous face. The nested
intersection is a single probability (extreme because it lies on all faces; a
convex combination decomposition would contradict max at some f_j). So it is
ergodic.
Nonatomicity: for any T-invariant nu, the coordinate a_i is uniform on
Z/m_i Z, because T cyclically permutes the level sets. If nu had a measurable
atom of mass c > 0, then for every i, up to null sets, it would lie in exactly
one a_i-level class, so c <= 1/m_i. Since m_i -> infinity, c = 0.
Standard: X is a compact metrizable subspace of countable product of finite
sets. Encode finite coordinates into disjoint ternary digit blocks {0, 2};
this gives a continuous injective map of X into a compact subset of [0, 1].
The image law under mu is nonatomic so its distribution function F is
continuous, and the quantile Q is a measurable inverse. This gives a
measurable isomorphism (X, mu) -> ([0, 1], Lebesgue).
Zero entropy: this is the important one. Take any partition P depending on
only FINITELY many coordinates: omega_{-r}, ..., omega_r and a_I for one level
I. The compatibility (i) determines all lower phases a_1, ..., a_{I-1} from
a_I. And a_j for any j >= I determines the boundaries of all level-j words
intersecting [-r, r].
Fix j >= I. The interval [-r, n-1+r] used in the n-step name of the
partition meets at most ceil((n + 2r)/m_j) + 1 level-j words. Each is one of
L_j words. So the number of possible n-step names is at most
m_j * L_j^{ceil((n+2r)/m_j) + 1},
which grows as m_j * exp(R_j n (1 + o(1))). The entropy rate of P is at most
R_j = (log L_j)/m_j. Taking first j -> infinity and then letting P range over
finite partitions of the phase extension gives h_mu(T) = 0. In fact this bound
holds for EVERY invariant measure, not just our chosen mu.
Finally, an arbitrary finite partition P (with A atoms) can be approximated by
a finite-coordinate partition Q with mismatch probability s arbitrarily small,
because cylinder sets generate the Borel sigma field. Chain rule:
H_n(P) <= H_n(Q) + H(P | Q),
and
H(P | Q) <= H_bin(s) + s log A.
Divide by n, take n -> infinity, then s -> 0. This shows h_mu(T, P) = 0 for
every finite P, so h_mu(T) = 0 for the phase extension.
---------------------------------------------------------------------------
ALIGNED WINDOWS AND THE WINDOW-ENTROPY STRUCTURE.
---------------------------------------------------------------------------
Fix once and for all an ergodic mu on X (from Proposition 11.1). For
x = (omega, a) in X, define the level-i word containing the origin:
V_i(x) in W_i, the string of length m_i beginning at the boundary
-a_i(x) that contains position 0.
And let A_i = {x in X : a_i(x) = 0} be the "level-i boundary at origin" set.
Define the CONDITIONAL MEASURE
mu_i := m_i * (mu restricted to A_i),
which is a probability (uniformizing the phase at origin).
The boundary-start law mu_i is the key object. It says: pick a point in X at a
level-i boundary, look at the words that begin at that boundary.
Define the advance function r_i(x, t) in {0, ..., m_i - 1}: it is ( - a_i(T^t x) )
mod m_i, the distance from t to the next level-i boundary at or after t.
Define, for D >= 2, the ALIGNED LIST
L_{i,D}(x, t) = ( V_i(T^{t + r_i(x,t) + j*m_i} x) )_{j = 0}^{D - 2}.
So L_{i,D}(x, t) is the list of the first D-1 full level-i words starting at
the first level-i boundary at or after t.
Lemma 11.2 (Aligned law). The law of L_{i,D}(x, 0) under mu is the same as the
law under mu_i of the first D-1 level-i words beginning at zero. Under mu_i,
the parent slot index
J := a_{i+2} / m_i
is uniform on {0, 1, ..., N_i N_{i+1} - 1}, and is independent of V_{i+2}: for
each such j and each v in W_{i+2},
mu_i(J = j, V_{i+2} = v) = mu_{i+2}(V_{i+2} = v) / (N_i N_{i+1}).
Proof. Invariance allows t = 0. Phase change: T^{-a_i} maps the level-i phase
class onto A_i, so mu_i captures the boundary-start law. Slot uniformity:
T^{j*m_i} permutes the sets {a_{i+2} = 0, V_{i+2} = v} cyclically, so
each has the same mu-mass. Multiply by m_i and use m_{i+2}/m_i = N_i N_{i+1}.
The uniformity (ii) is what makes "every parent word conditioned" turn into
"every offset sampled uniformly" in the entropy computation.
---------------------------------------------------------------------------
WINDOW ENTROPY IS ALMOST MAXIMAL.
---------------------------------------------------------------------------
Two elementary entropy bounds we will use.
Bound 1. If a law on a finite set has every atom of mass at most M, then its
entropy is at least log_2 (1/M).
Bound 2. If two laws P, Q on an A-point set satisfy TV(P, Q) <= s, then
H(P) >= H(Q) - 1 - s log_2 A.
Proof of Bound 2: couple them with mismatch probability <= s, record the
mismatch flag, chain rule.
Proposition 11.3 (Window entropy). For all sufficiently large i, for every
power of two D with 2 <= D <= D_i^*, and every integer t:
H_mu( L_{i,D}(., t) ) >= .996 * (D - 1) * ell_i.
Here ell_i = log_2 L_i, and the threshold on i is independent of D, t, and the
specific invariant mu on X.
Proof sketch. Work under mu_i (the boundary-start law). Write q = D - 1. Split
the uniform parent slot J as
J = h*N_i + u, 0 <= h < N_{i+1}, 0 <= u < N_i.
Conditional on the parent v = V_{i+2} and the parent-internal offset u, the
slot h is uniform on {0, ..., N_{i+1}-1}. Call u ADMISSIBLE if u + D_i^* <= N_i.
The non-admissible fraction is beta_i := (D_i^* - 1)/N_i, tending to 0.
For fixed v and admissible u:
- By (W4) at level i+1, the law of the D_i^*-window at offset u is within
10^{-4} in TV of the law of the same window in a uniform word of W_{i+1}.
- By (W3) at level i, that law is within a further 10^{-4} of Q_{i,t}, whose
atom at any prescribed q-position tuple is at most
M_i = 2 * (2^{2 K_i} / L_i)^q.
Restriction to the first q entries cannot increase TV distance. So by Bound 2
with s = 2 * 10^{-4} and alphabet size L_i^q, the actual q-list conditionally
on (v, u) has entropy at least
q * ell_i - 2 K_i q - 1 - (1 + s q ell_i) = q * ((1 - s) ell_i - 2 K_i) - 2.
Mixing over (v, u) preserves this by concavity of entropy; the non-admissible
fraction beta_i only reduces this by a factor (1 - beta_i). So
H_mu( L_{i,D}(., t) ) >= (1 - beta_i) * ( q((1-s) ell_i - 2 K_i) - 2 ).
Now beta_i -> 0, and 2 K_i / ell_i -> 0, and 2 / ell_i -> 0. Once each is at
most .0005, dividing by q*ell_i, with q >= 1:
(1 - .0005) * (1 - .0002 - .0005 - .0005) > .996.
So the ratio H / (q ell_i) is at least .996, proving the proposition.
Note: only two consecutive levels i and i+1 enter this argument. The TV errors
do NOT accumulate over levels; the total variation cost is exactly 2 * 10^{-4}.
---------------------------------------------------------------------------
THE GRAPH RELATION ACROSS TWO WINDOWS.
---------------------------------------------------------------------------
Two legal level-i words u, v in W_i pass the GRAPH TEST at level i if
[u, v]_i = 0,
using the symplectic labels. Now consider a fixed point x in X and its two
aligned windows: the PAST window L_{i,D}(x, -n) and the FUTURE window
L_{i,D}(x, 0). Both are lists of D-1 words. Question: for how many j in
{0, ..., D-2} does the pair (past entry j, future entry j) pass the graph
test?
Lemma 11.4 (Graph windows). Let 2 <= D <= D_i^* be a power of two and set
n = D * m_i. For all sufficiently large i, with mu-probability at least
1 - alpha_{i,D}, alpha_{i,D} := (2D - 2)/H_i,
ALL corresponding pairs of past and future entries pass the graph test at
level i simultaneously. In particular, sup_{D <= D_i^*} alpha_{i,D} -> 0.
Proof sketch. Since n is a multiple of m_i, both windows advance by the same
q_0 = r_i(x, 0). Entry starts are
past: -n + q_0 + j*m_i
future: q_0 + j*m_i
for 0 <= j <= D-2. Corresponding entries are separated by exactly D level-i
positions. The first start through the last start SPAN 2D - 1 level-i
positions (including the unused middle block between the two windows).
Now, from the aligned law (Lemma 11.2), the first boundary position of the
past window is uniform mod H_i (because the parent-internal position is
uniform mod N_i, and H_i divides N_i). Since 2D-1 <= H_i (by D <= D_i^* and
H_i >> D_i^*), exactly 2D - 2 residues have a span that crosses some H_i-group
boundary. Every OTHER residue has all 2D-1 aligned positions inside a SINGLE
group.
But (W1) is exactly the statement that within each group, positions separated
by a power of two satisfy [u, v]_i = 0. Since D is a power of two, this
applies. So on every residue NOT crossing a group boundary, ALL D-1 pairs
pass simultaneously.
The bad fraction is exactly (2D-2)/H_i, and 2D_i^* / H_i -> 0.
The graph lemma is what makes the symbolic side "rigid": nearly all pairs of
windows pass the graph test, forced by (W1). And the smooth switch, by
contrast, forces independence of the two windows, so the smooth side can only
pass the graph test rarely.
That is the setup. Now the contradiction.
===============================================================================
12. THE OBSTRUCTION
===============================================================================
The punchline: on the smooth side, matching past and future lists are
INDEPENDENT (conditional on the cell and one entry), so the fraction of
graph-test passes is at most about 62%. On the symbolic side, they are
FORCED to pass almost always by (W1). Contradiction.
This section pins down the comparison.
---------------------------------------------------------------------------
FIRST: FINDING CHEAP SCALES ON THE SYMBOLIC SIDE.
---------------------------------------------------------------------------
Lemma 12.1 (Cheap scales). Let (B_k)_{k>=1} be any POSITIVE sequence with
sum_k B_k^{16} < infinity. Then there exist a subsequence of levels i -> infinity
and, at each such i, a dyadic D with
2^{ceil(K_i/16)} <= D <= 2^{floor(K_i/8)},
such that, with n = m_i * D:
B_{log n} / R_i -> 0 along the subsequence,
where R_i = ell_i / m_i.
In particular, the smooth partition at scale log n has
log #D_{log n} = o( (D - 1) ell_i ),
for every fixed constant in the partition-size bound.
Proof. Set b_i' = log_2 m_i, then b_{i+1}' = b_i' + K_i. Consider the integer
sets
J_i := { b_i' + j : ceil(K_i/16) <= j <= floor(K_i/8) }.
These sets are disjoint (each covers a different range of dyadic scales),
and for large i they have at least K_i / 32 integers each.
Assume for contradiction that B_{log n} / R_i >= c > 0 for all sufficiently
late tests. Then
sum_k B_k^{16} >= c^{16} * sum_i |J_i| * R_i^{16}
>= c^{16} * (1/32) * sum_i K_i R_i^{16}
= infinity,
because sum_i K_i R_i^{16} = infinity (Lemma 10.3d). But sum_k B_k^{16} is
finite. Contradiction.
So there are scales where conditioning on the smooth partition cell loses at
most a vanishing fraction of the window entropy.
---------------------------------------------------------------------------
SECOND: THE SMOOTH MODEL, AND THE UPPER BOUND FROM INDEPENDENCE.
---------------------------------------------------------------------------
Suppose for a contradiction that (X, mu, T) has a smooth positive-volume model
(M, nu, S). Its dimension is positive because a nonatomic probability cannot
live on a zero-dimensional compact manifold (which is finite).
Apply the switch theorem (section 6 plus sections 7-9) with tolerance eta = .01.
All the smooth data, the sequence B_k, and the constant C are now fixed.
Use cheap scales Lemma 12.1 for this sequence. So there is a subsequence of
i, D, n with n = m_i D and B_{log n} / R_i -> 0, and
log #D_{log n} = o( (D-1) ell_i ).
Identify X, via the smooth isomorphism, with M. So mu is carried to nu.
At each such test scale, let C be the cell of D_k sampled in the pair law
rho_k, and (x, y) be the retained pair.
Define two lists:
U := L_{i,D}(y, -n) (past window, aligned under y)
V := L_{i,D}(x, 0) (future window, aligned under x)
q := D - 1
And define the passing statistic
Z(x, y) := (1/q) * sum_{j=0}^{q-1} 1_{ [U_j, V_j]_i = 0 },
the FRACTION of corresponding entries that pass the level-i graph test.
For each fixed cell and each fixed j, the two entries U_j and V_j are drawn
INDEPENDENTLY: U_j from y's past law, V_j from x's future law, conditional on
(x, y) sharing cell C. Independence is EXACT, because rho_k conditioned on C
is a product of two copies of (nu|_C).
Upper bound on Z: by the window-entropy prop and independent marginals,
H(U), H(V) >= .996 * q ell_i.
Conditioning each vector on the sampled cell loses at most H(C) <= log #D_{log n}
= o(q ell_i). So conditional subadditivity:
(1/q) sum_{j=0}^{q-1} H(U_j | C) >= .99 ell_i,
(1/q) sum_{j=0}^{q-1} H(V_j | C) >= .99 ell_i.
Apply the graph-independent lemma (below) with index xi = (C, j), j uniform
independent of the cell. So conditional on the full index, U_j and V_j are
independent samples from two high-entropy laws on nonzero vectors.
LEMMA 12.2 (Graph independence). Let L = 2^{2s} - 1 (so L+1 is a power of 2),
equipped with a nondegenerate alternating form. Suppose a family of pairs
of laws (p_xi, q_xi) indexed by a probability space satisfies
E H(p_xi) >= .99 log_2 L, E H(q_xi) >= .99 log_2 L.
Then for all sufficiently large L, the average over xi of the probability
that independent samples from p_xi and q_xi pass [u, v] = 0 is at most .62.
Sketch proof. For a law p, let A_p = {u : p(u) > L^{-.8}} (the "heavy"
points), and theta_p = p(A_p). Heavy entropy bound:
H(p) <= 1 + .8 * theta_p * log L + (1 - theta_p) * log L
= 1 + (1 - .2 theta_p) * log L.
So E theta_p <= .055 when log L >= 1000.
Matrix A_{uv} = (-1)^{[u,v]}: on the FULL vector space, its rows are
orthogonal (nondegeneracy makes v -> [u+w, v] a nonzero functional for
u != w, so the sum over v vanishes). Each row has squared norm L+1. So the
full operator norm is sqrt(L+1), and the nonzero-block has no larger norm.
Light subprobability p_l = p * 1_{A_p^c} satisfies ||p_l||_2^2 <= L^{-.8}.
So |p_l * A * q_l| <= sqrt(L+1) * L^{-.8} =: beta_L, and beta_L <= .02 for
large L.
The passing mass of the light-light pair is at most (1 + beta_L)/2.
Plus the heavy-heavy and heavy-light parts contribute at most
E theta_p + E theta_q <= .055 + .055.
Sum:
.055 + .055 + (1 + .02)/2 = .62.
The diagonal of A is included (u = v has [u, v] = 0), so no separate
exclusion of diagonal is needed.
Conclusion:
E_rho Z <= .62. (**)
---------------------------------------------------------------------------
THIRD: THE LOWER BOUND FROM SHADOWING.
---------------------------------------------------------------------------
Now we use the shadowing property of the switch. Recall from the geometric
construction (section 9): the retained set G_k has rho_k-mass >= 1 - eta, and
for (x, y) in G_k the output z satisfies: for 0 <= t <= n, S^t z is close to
S^t x; for -n <= t <= 0, S^t z is close to S^t y; and the closeness tends to
zero UNIFORMLY as k -> infinity.
We want to translate "geometric closeness" into "symbolic agreement". Use a
compact-cores argument.
Lemma 12.3 (Symbol matching). For all sufficiently large dyadic times n in
the smooth switching estimate, the retained pairs (x, y) in G_k for which
# { 0 <= t < n : f(S^t z) != f(S^t x) } < d_* n / 100
# { -n <= t < 0 : f(S^t z) != f(S^t y) } < d_* n / 100
have original rho-mass at least .96. Here f: M -> A is the transported symbol
at time 0.
Sketch proof. Inner regularity of the smooth probability nu gives compact
sets K_a contained in f^{-1}(a), for a in A, whose union has mass at least
1 - kappa of nu, with kappa << d_* / 20000. Measurable atoms are replaced by
Borel representatives modulo null sets. Each core is compact, and the
distance between any two nonempty distinct cores is positive (uniform
separation). Uniform shadowing error at large scales is smaller than this
separation, so two compared iterates both in cores have the SAME symbol.
Write G = G_k and sigma = z_* (rho |_G). The switch gives sigma <= 2 nu, and
by pushing forward invariance, (S^t)_* sigma <= 2 nu for every t. The x and
y time marginals under rho are nu. So at each future comparison time, the
rho-mass within G where either x or z misses the cores is at most 3 kappa.
Same on the past half. Integral of mismatch fraction on each half <= 3 kappa.
Markov at threshold d_* / 100 bounds the failure mass on each half by
600 kappa / d_* < .03. Since rho(G) >= 1 - eta >= .99, joint success mass
>= .96. All estimates are in the ORIGINAL law, without renormalizing.
Now the punch line: on the "matching" event, the phases agree.
If a_i(z) != a_i(x), then on each of the D-1 full level-i words beginning at
x's first boundary in [0, n), the symbol string of z is a window of length m_i
cut at a NONZERO offset across two consecutive legal level-i words. By
recognition (Lemma 10.2), such a window differs from any legal word in at
least d_* m_i positions. Over D-1 disjoint intervals that is
>= d_* (D-1) m_i = d_* n * (D-1)/D >= d_* n / 2
mismatches, contradicting the < d_* n / 100 mismatch budget from the matching
event.
So
a_i(z) = a_i(x) = a_i(y). (phase recovery)
Here a_i(S^{-n} u) = a_i(u) because n is a multiple of m_i.
Then the two aligned entries U_j, V_j occupy, in the symbolic orbit of z, the
positions -n + q_0 + j m_i and q_0 + j m_i. Their DIFFERENCE is exactly
n = D m_i, and their offset from z's boundaries is uniform. And on the
matching event, the actual labels match those of z at those positions, EXCEPT
for at most n/100 * (1/2 + 1/2) = n / 100 positions total (combining both
halves), so at most D / 100 entries differ on each side.
Finally, apply the graph-window lemma (Lemma 11.4) for the ACTUAL symbolic
orbit of z: with mu-probability at least 1 - alpha_{i,D} (which is close to 1
for late tests), all D-1 pairs of z's window entries pass the graph test.
Push this through sigma = z_* (rho |_G) <= 2 nu: the mass of pairs with z in
the graph-exception set E_{i,D} is at most 2 alpha_{i,D}. Choose a late test
with 2 alpha_{i,D} <= .01.
On the intersection of:
- the matching event (mass >= .96) minus graph-exception (mass <= .01):
at least .95 of the pair space,
- and for those pairs, removing at most D/100 wrong entries per side,
if the only wrong entries are fewer than D/100 on each side, then
Z(x, y) = (1/(D-1)) * #{j : [U_j, V_j]_i = 0}
>= (D-1 - 2D/100) / (D - 1)
>= 1 - (2D/100) / (D - 1)
>= .96.
Z is nonnegative everywhere else, so
E_rho Z >= .95 * .96 >= .912. (***)
---------------------------------------------------------------------------
FOURTH: THE CONTRADICTION.
---------------------------------------------------------------------------
(**) says E_rho Z <= .62. (***) says E_rho Z >= .912.
Both are statements about the SAME quantity E_rho Z, computed under the same
original pair law rho, at the same set of test scales.
The two inequalities are incompatible. Contradiction.
Since the only assumption was the existence of a smooth positive-volume model,
there is no such model.
===============================================================================
13. WHAT THIS PAPER DOES NOT CLAIM
===============================================================================
To avoid confusion with related results, here is a precise list of what this
paper does and does not say.
CLAIMED: There exists ONE ergodic invertible mpt (X, mu, T) of a standard
nonatomic probability space with h_mu(T) = 0 such that (X, mu, T) has NO
smooth positive-volume model on ANY finite-dimensional compact C^infinity
manifold, regardless of boundary and orientability, with ANY smooth positive
density and ANY C^infinity volume-preserving diffeomorphism.
NOT CLAIMED:
(1) That every finite-entropy ergodic mpt fails to have a smooth
positive-volume model. This paper gives ONE counterexample. Plenty of
finite-entropy systems DO have smooth models (e.g. Bernoulli shifts
are isomorphic to diffeomorphisms of the two-torus, via
Lind-Thouvenot).
(2) That full-support invariant measures cannot occur as invariant measures
of smooth diffeomorphisms. Quas-Soo 2016 shows they can, and their
measure can be singular. Only the SMOOTH POSITIVE DENSITY restriction
is being used here.
(3) That our system fails to have a model with a merely absolutely-continuous
invariant measure. Foreman-Weiss 2022 treats measures that are
equivalent to volume but not necessarily smooth positive density.
(4) That our system fails to have a Lipschitz model, or a homeomorphism
model. Katok-Thouvenot 1997 treats Lipschitz and slow-entropy
obstructions for Z^2 actions; their obstruction does not apply to the
single-transformation problem. Lind-Thouvenot 1977 gives homeomorphism
models for every finite-entropy system on the two-torus with Lebesgue
measure.
(5) That smooth systems are hard to classify. Anti-classification results
(Foreman-Weiss, Kunde, Gerber-Kunde) say the measure-isomorphism relation
is not Borel on certain families of smooth systems. That is a statement
about equivalence classes, not about individual realizations.
(6) That our zero-entropy system has any unusual properties beyond the
specific counterexample. It is a construction from word families. It
is ergodic, invertible, zero-entropy, standard nonatomic. That is all.
(7) That the proof is sharp. The proof leaves gaps like "eta < 0.01" and
"K_1 >= 256". The exact numerical constants are not optimized. A
sharper version would presumably tighten these, and might improve
the .62 vs .912 gap. But those are cosmetic changes.
===============================================================================
14. GLOSSARY OF EVERY SYMBOL
===============================================================================
Alphabetical and structural, in order of appearance.
STANDARD SETS AND SPACES
R, Z, N real, integer, natural numbers.
F_2 two-element field {0, 1}, addition mod 2.
F_2^{2s} vector space of dimension 2s over F_2.
A alphabet of the symbolic system. Here A = A_1 = F_2^{2s_1}\{0}.
A_i alphabet used at level i (nonzero vectors of F_2^{2s_i}).
[0,1] unit interval.
MEASURES AND TRANSFORMATIONS
(X, mu) probability space.
T invertible measure-preserving transformation of (X, mu).
(M, nu) compact smooth manifold with smooth positive probability
density nu.
S C^infinity diffeomorphism of M, preserving nu.
eta fixed tolerance used in the switch theorem (e.g. .01).
tau small wall margin inside product cells (e.g. .001).
kappa tolerance for core-exception mass in the symbol-matching lemma.
ENTROPY
H(Y) Shannon entropy of a finite random variable Y.
H(Y | Z) conditional entropy of Y given Z.
h_mu(T) Kolmogorov-Sinai entropy of T with respect to mu.
H_bin(s) binary entropy function -s log_2 s - (1-s) log_2(1-s).
TV(P,Q) total variation distance (1/2) sum_a |P(a) - Q(a)|.
MANIFOLD AND DERIVATIVE
d dimension of M.
T_x M tangent space at x.
D S_x derivative of S at x, a linear map T_x M -> T_{S x} M.
bigwedge^p p-th exterior power of a linear map.
J_p(A | E) expansion factor of A on p-dimensional subspace E (up to
unit simple p-vector).
Lambda sum of the fast rates.
Omega minus the sum of the slow rates (note Omega is generally
positive; it is the "backward rate" of the slow group).
gamma threshold rate separating fast from slow.
p (in section 8) dimension of the fast part.
FRAME BUNDLE AND TRIANGULAR COCYCLE
F M orthonormal frame bundle.
e frame: an isometry R^d -> T_x M.
R(e) upper-triangular positive-diagonal factor in the QR
factorization D S_x o e = tilde S(e) o R(e).
tilde S(e) new frame at S x after Gram-Schmidt.
g_j(e) ln(R(e)_{jj}), the j-th diagonal log.
A_h block average (1/2^h) sum_{t<2^h} Y_t.
Z_h normalized maximum chord deviation at scale 2^h.
v_l E(A_{l-1}^2) - E(A_l^2), the "variance increment" at level l.
SCALES AT DYADIC TIME
n 2^k, dyadic time in the smooth switch.
b_k the small positive sequence with b_k >= 2^{-k/10} and
sum b_k^2 < infinity.
B_k b_k^{1/8}. sum B_k^16 < infinity.
B generic notation for B_k in sections 6 and 9.
g b_k^{1/2}. Rate separation margin.
L ceil(b_k^{1/4} n). Short comparison time.
h_0 e^{-B n}. Coarse bin side.
r e^{-2 B n}. Product cell side (much smaller than h_0).
delta e^{-3 B n/2}. Contraction uniqueness radius.
Q_k e^{C g n}. Unweighted inverse bound for the linearized
boundary problem.
PARTITIONS AND CELLS
D_k partition of M into cells at scale k.
Q, Q' coarse bin (uniform coordinate cube of side h_0).
product cell parallelepiped spanned by E_Q^s and E_Q^f with sides r.
R_k retained set in M.
G_k retained set in M x M.
rho_k cell-weighted pair law on M x M.
z, w complementary switch outputs from a retained pair (x, y).
F_x, F_y future endpoint projections.
P_x, P_y past endpoint projections.
chi_U coordinate localization map in the coarse bin U.
U_x, V_x coarse bins of S^n x and S^{-n} x respectively.
U_y, V_y coarse bins of S^n y and S^{-n} y.
JACOBIAN MINORS
H_{ab}(u) (F_a(u), P_b(u)), a,b in {x,y}.
A_x, A_y leading p-form minors of DS^n at x and at y.
B_x, B_y leading (d-p)-form minors of DS^{-n} at x and at y.
WORD CONSTRUCTION
s_i half the vector dimension at level i (so F_2^{2s_i}).
K_i floor(sqrt(ell_i)).
ell_i log_2 L_i.
L_i 2^{2s_i} - 1: number of words in W_i.
W_i family of L_i legal level-i words.
N_i 2^{K_i}: number of level-i words in a level-(i+1) word.
H_i 2^{floor(K_i/2)}: number of "groups" used in (W1).
D_i^* 2^{floor(K_i/8)}: bound on window lengths and lags.
m_i length (in level-1 symbols) of a level-i word.
m_1 = 1, m_{i+1} = m_i N_i.
eps_i 2^{-i-2}. Distance slack at level i.
vartheta 10^{-4}. TV tolerance in (W3) and (W4).
[.,.]_i alternating form on F_2^{2s_i}.
c_i(w) constituent list of w in W_{i+1}, of length N_i.
d_H Hamming distance between equal-length words.
d_i prod_{j<i} (1 - eps_j).
d_* lim d_i = prod_{j>=1} (1 - eps_j) >= 3/4.
R_i ell_i / m_i. Entropy per level-1 symbol.
PROCESS AND WINDOWS
Omega A^Z x prod_i Z/m_i Z.
X closed T-invariant subset of Omega (the phase extension).
(omega, a) a point: a bi-infinite symbol sequence and all alignment
phases.
a_i the level-i phase coordinate (position of origin within the
level-i word).
T the shift on X.
V_i(x) the level-i word containing the origin at x.
A_i {x : a_i(x) = 0}.
mu_i m_i * (mu restricted to A_i). Boundary-start law.
r_i(x, t) advance to the first level-i boundary at or after t.
L_{i,D}(x,t) aligned list of D-1 full level-i words starting at the first
level-i boundary at or after t.
q D - 1.
alpha_{i,D} (2D-2)/H_i. Graph-window exception mass.
OBSTRUCTION
J_i interval {b_i' + j : ceil(K_i/16) <= j <= floor(K_i/8)} of
dyadic scales contributed by level i.
U, V past and future aligned windows of a retained pair.
Z(x, y) passing statistic: fraction of corresponding entries passing
graph test.
E_{i,D} graph-exception set (transported from X to M).
theta_p, A_p heavy-mass and heavy-set of a probability p.
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CLOSING REMARKS
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The proof has two halves that meet at a single number.
On the smooth side, the switching construction (sections 6 through 9) yields
a family of test scales with sum_k B_k^{16} < infinity, and gives the pair
law rho_k exact geometric shadowing plus a switch with unnormalized volume
output bounded by 2 nu. When the corresponding symbolic observations are
pushed through the switch, independence (from the product structure of rho_k
conditioned on a cell) forces E_rho Z <= .62.
On the symbolic side, the word construction (sections 10 through 12) produces
an ergodic zero-entropy system whose short window entropy at every level is
nearly maximal, and whose graph-window lemma forces E_rho Z >= .912 on the
same pair law rho_k at the same scales.
The two numbers conflict. No smooth positive-volume model can exist.
The subtle points, if you remember nothing else:
- the RATE (square-summable dyadic errors) on the smooth side has a
definitive adversary: the SYMBOLIC rate is not square-summable
(sum K_i R_i^{16} diverges). This is why the smooth partition costs
become negligible compared to symbolic entropy at infinitely many scales.
- the DENSITY (smooth positive) is what gives the switch its 2 nu output
bound. A merely full-support or merely absolutely continuous measure
would let the pair map concentrate. This is the exact sense in which
"smooth positive-volume" is used.
- the CONJUGACY does not need to be continuous. Measurability is enough.
Everything happens on invariant conull sets. The smooth side uses
measurable facts about the density; the symbolic side uses combinatorics
of the word structure; the two sides communicate only through the
measurable notation.
The construction deliberately uses a single example of a symbolic system
(fixed once, then all smooth models of any dimension are ruled out). The
reverse order (choose a smooth manifold first, then pick a counterexample)
would be strictly weaker.
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